Download ± PSS 6.1 Work and Kinetic Energy

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± PSS 6.1 Work and Kinetic Energy
Learning Goal:
To practice Problem­Solving Strategy 6.1 Work and Kinetic Energy.
Your cat “Ms.” (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined 19.0 ∘ above
the horizontal. Since the poor cat can’t get any traction on the ramp, you push her up the entire length of the ramp by
exerting a constant 52.0 N force parallel to the ramp. If Ms. is moving at 2.00 m/s at the bottom of the ramp, what is her
speed when she reaches the top of the incline?
Problem­Solving Strategy: Work and kinetic energy
IDENTIFY the relevant concepts: The work–energy theorem, Wtot = K2 − K1 , is extremely useful when you want to relate a body’s speed v 1 at one point
in its motion to its speed v 2 at a different point. (It’s less useful for problems that involve time.)
SET UP the problem using the following steps:
1. Choose the initial and final positions of the body, and draw a free­body diagram showing all the forces that act
on the body.
2. Choose a coordinate system. (If the motion is along a straight line, it’s usually easiest to have both the initial
and final positions lie along the x axis.)
3. List the unknown and known quantities; decide which ones are your target variables.
EXECUTE the solution as follows:
1. Calculate the work W done by each force. If the force is constant and the displacement is a straight line, use W = F s cos ϕ
or W
⃗ = F ⋅ s ⃗ ⃗ . Recall that W must be positive if F has a component in the direction of the
⃗ displacement, negative if F has a component opposite to the displacement, and zero if the force and the
displacement are perpendicular.
2. Add the amounts of work done by each force to find the total work Wtot . Sometimes it’s easier to find the
work done by the net force acting on the body; this value is also equal to Wtot .
3. Write expressions for the initial and final kinetic energies, K1 and K2 . Note that kinetic energy involves
mass, not weight.
4. Finally, use Wtot = K2 − K1 to solve for the target variable. Remember that K2 − K1 is the final kinetic
energy minus the initial kinetic energy, never the other way around.
EVALUATE your answer: Check whether your answer makes physical sense.
IDENTIFY the relevant concepts
The target variable in this problem is the cat's speed at the top of the ramp. Because you are given the cat's speed at the
bottom of the ramp, you can use the work–energy theorem to relate the cat's speed at the two different points in her motion.
SET UP the problem using the following steps
Part A
The initial and final positions of the cat are at the bottom and the top of the ramp, respectively. Since the forces acting
on the cat are constant, a free­body diagram can be drawn at any arbitrary position in her motion.
Use the diagram below to draw a free­body diagram showing all the forces acting on the cat. The cat is represented by
the black dot.
Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length
of the vectors will not be graded.
ANSWER:
Correct
Now, make your own sketch of a free­body diagram for this problem. If you haven't done so yet, make sure the
magnitudes of vectors in the diagram are physically reasonable.
Part B
Now, draw the most convenient coordinate system for this problem.
The orientation of the vectors will be graded. The location and length of the vectors will not be graded.
ANSWER:
Correct
Since the motion is along a straight line, it’s usually easiest to have both the initial and final positions lie along the
x axis. In this way, any force perpendicular to that axis will do no work on the body.
Now, make a list of the known and unknown variables, and complete your sketch like this:
EXECUTE the solution as follows
Part C
What is the cat's speed v 2 when she reaches the top of the incline?
Express your answer in meters per second to three significant figures.
Hint 1. How to use the work–energy theorem
To find a body's speed at the initial and final positions of its motion using energy considerations, calculate the
total work done on the body and the body's kinetic energy. As explained in the strategy, you can calculate the
total work Wtot done on the body by adding the amounts of work done by each force acting on the body.
Alternatively, you can calculate the vector sum of the forces (the net force) and then find the work done by the
net force. In this tutorial, we'll use the second method.
Hint 2. Find the total work done on the cat
Calculate the total work Wtot done on the cat when she reaches the top of the ramp.
Express your answer in joules to three significant figures.
Hint 1. Find the x component of the net force acting on the cat
Find Fx,net , the component of the net force acting on the cat parallel to the displacement. Recall that if a
force does not point directly in the x or y direction, you will have to find its x and y components.
Express your answer in newtons to three significant figures.
Hint 1. Trigonometric relations
In a right triangle such as the one shown in the figure below, the sides of the triangle are related to
the angle ϕ by the following equations:
and b
a = c cos ϕ
= c sin ϕ
ANSWER:
F x,net
= 24.9 N ANSWER:
Wtot
= 49.8 J Hint 3. Find the cat's initial kinetic energy
If the cat takes a running start so that she is moving at 2.00 m/s at the bottom of the ramp, what is her kinetic
energy K1 at that point?
Express your answer in joules to three significant figures.
Hint 1. Kinetic energy
The kinetic energy of a body of mass m that moves with speed v is defined as
K=
1
2
2
mv
ANSWER:
K1
= 17.0 J Hint 4. Find the cat's final kinetic energy
Use the results you found in the previous hints and apply the work–energy theorem to find the cat's final kinetic
energy, K2 .
Express your answer in joules to three significant figures.
ANSWER:
K2
= 66.8 J ANSWER:
v2
= 3.96 m/s Correct
EVALUATE your answer
Part D
This problem can also be done without the work–energy theorem.
Since the forces acting on the cat are constant, the cat is moving up the ramp with a constant acceleration. Therefore,
if you know her acceleration, you can use the equations of motion for constant acceleration to find her final speed. Use
Newton's 2nd law to find the magnitude a of the cat's acceleration.
Express your answer in meters per second squared to three significant figures.
Hint 1. Find the net force
Find Fx,net , the component of the net force acting on the cat parallel to the displacement. Note that you may
have already found this information while working through Part C.
Express your answer in newtons to three significant figures.
ANSWER:
F x,net
= 24.9 N ANSWER:
= 2.93 m/s2 a
Correct
Now, use the equation v 22 = v 21 + 2as, where s is the length of the ramp, to find the cat's final speed v 2 . As you
can see, you get the same result as you obtained in Part C using the work–energy approach, but there you
avoided the intermediate step of finding the acceleration. When a problem can be done by two different methods,
doing it by both methods is a very good way to check your work!
Problem 6.36
Part A
An unusual spring has a restoring force of magnitude F = (2.00 N/m)x + (1.00 N/m2)x 2, where x is the stretch of the
spring from its equilibrium length. A 3.00­kg object is attached to this spring and released from rest after stretching the
spring 1.30 m. If the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to
its equilibrium length?
ANSWER:
1.78 m/s
0.890 m/s
2.03 m/s
1.27 m/s
1.52 m/s
Correct
Problem 7.62
A 3.00­kg block is connected to two ideal horizontal springs having force constants k1 = 28.0 N/cm and k2 = 19.0 N/cm
(the figure ). The system is initially in equilibrium on a horizontal,
frictionless surface. The block is now pushed 15.0 cm to the right
and released from rest.
Part A
What is the maximum speed of the block?
ANSWER:
v max
= 5.94 m/s Correct
Part B
Where in the motion does the maximum speed occur?
ANSWER:
3759 Character(s) remaining
Þar sem þeir voru upphaflega í jafnvægi. Graded, see 'My Answers' for details
Part C
What is the maximum compression of spring 1?
ANSWER:
x1
max
= 15.0 cm Correct
Shooting a Block up an Incline
A block of mass m is placed in a smooth­bored spring gun at the bottom of the incline so that it compresses the spring by
an amount xc . The spring has spring constant k. The incline makes an angle θ with the horizontal and the coefficient of
kinetic friction between the block and the incline is μ. The block is released, exits the muzzle of the gun, and slides up an
incline a total distance L.
Part A
Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside
the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance xc
while inside of the gun). Use g for the magnitude of acceleration due to gravity.
Express the distance L in terms of xc , k, m, g, μ, and θ.
Hint 1. How to approach the problem
This is an example of a problem that would be very difficult using only Newton's laws and calculus. Instead, use
the Work­Energy Theorem: Ef inal − Einitial = Wext , where Ef inal is the final energy, Einitial is the initial
energy, and Wext is the work done on the system by external forces. Let the gravitational potential energy be
zero before the spring is released. Then, Einitial is the potential energy due to the spring, Ef inal is the potential
energy due to gravity, and Wext is the work done by friction. Once you've set up this equation completely, solve
for L.
Hint 2. Find the initial energy of the block
Find the initial energy Einitial of the block. Take the gravitational potential energy to be zero before the spring is
released.
Express your answer in terms of parameters given in the problem introduction.
Hint 1. Potential energy of a compressed spring
Recall that the potential energy U of a spring with spring constant k compressed a distance x is 2
U = (1/2)kx .
ANSWER:
Einitial
= 1
2
kx c
2
Hint 3. Find the work done by friction
Find Wf riction , the work done by friction on the block.
Express Wf riction in terms of L, m, g, μ, and θ.
Hint 1. How to compute work
The work done by a force acting along the direction of motion of an object is equal to the magnitude of
the force times the distance over which the object moves. Work is negative if the force directly opposes
the motion.
ANSWER:
Wf riction
= −Lmgcos(θ)μ
Hint 4. Find the final energy of the block
Find an expression for the final energy Ef inal of the block (the energy when it has traveled a distance L up the
incline). Assume that the gravitational potential energy of the block is zero before the spring is released and that
the block moves a distance xc inside of the gun.
Your answer should contain L and xc .
Hint 1. What form does the energy take?
When the block stops sliding up the ramp, all of its energy is in the form of gravitational potential energy.
ANSWER:
Ef inal
= mg(L + xc )sin(θ)
ANSWER:
1
L
= 2
2
kx c −mgsin(θ)x c
mg(sin(θ)+μcos(θ))
Correct
Problem 7.70
A small block with mass 0.0400 kg slides in a vertical circle of radius 0.575 m on the inside of a circular track. During one
of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force
exerted on the block by the track has magnitude 4.00 N . In this same revolution, when the block reaches the top of its
path, point B, the magnitude of the normal force exerted on the block has magnitude 0.660 N .
Part A
How much work was done on the block by friction during the motion of the block from point A to point B?
Express your answer with the appropriate units.
ANSWER:
Wf riction
= ­0.283 J
Correct
Score Summary:
Your score on this assignment is 101%.
You received 5.04 out of a possible total of 5 points.